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Why astronauts during launch fly faster than jets but experience less Gs?
The NASA Space Shuttle used to lunch astronauts into spaces at high speeds, supersonic speeds. When you google whats the Gs force they experienced while departing or simply “Gs forces astronauts” You get 3gs. I watched an interview of an astronaut he confirmed it, it was about 3Gs. Lets say 4Gs. Why is it that even though they were experiencing higher than a jet aircraft speeds they were only experiencing 3-4gs, while jets fly much slower but experience many more Gs?
I think it has to be with the rate of acceleration? Like how fast the velocity vector is changing, correct? In magnitude or in direction or both (maybe)?
So would you experience the same amount of Gs if moving in a straight line either vertically or horizontally (assuming both have the same rate of acceleration change)?  Or no Gs at all?
Maybe Im confusing inertia like when you accelerate (or break) in your car in the freeway your body either move forward or backwards, right? But if thats the case, why do astronauts experience Gs at all (assuming they are moving straight up, is it because they are not?)?
*Also, I do understand how Gs forces are formed while flying in a turn, less vertical lift more total lift needed, then more L to W ratio. But I guess Im confused whats going on while moving in a straight path.

Basically you are asking for a complete  explanation of the concept of G-loading, and a specific question about the G-load felt on the space shuttle vs on airplanes?  That:s pretty broad.

You are confusing velocity and acceleration. Flight speed is velocity (meters persecond, miles per hour, knots, etc) while "Gs" are acceleration (meters per second per second, ).

I think it has to be with the rate of acceleration? Like how fast the velocity vector is changing, correct? In magnitude or in direction or both (maybe)?

You kind of have the right idea there in the second part of your statement. Acceleration is the rate at which the velocity vector is changing. The change in rate of acceleration is something else completely called a jerk/jolt.
Imagine two cars at a stop light. When the light turns green, one car floors the gas for a couple seconds and reaches 40 miles per hour. When that happens, you feel slammed into the back of your seat (higher Gs). The other car presses the gas halfway but holds it there for a minute and reaches 200 mph. With this car, you feel much less force pushing you back in your seat (lower Gs). The first car had a high acceleration but lower speed while the second car had a lower acceleration and a higher speed. The first car is like the fighter jet and the second is like the rocket.

Astronauts experience a G force due to the acceleration as the space craft gains speed. There is very little change in direction.
Pilots also experience a small amount of G force due to acceleration when they gain speed, but they experience far more g Forces when changing direction. The G force is experienced when the vector changes. When accelerating the vector gets longer. When turning the vector's orientation changes. Very high G force is only encountered in aircraft during very tight turns at high speed when the vectors are changing very rapidly.
Imaging a 3 meter long stick. Add 1 meter to the end of it making it 4 meter long. This a change of 1 meter. Now take the 3 meter stick and rotate it 180 degrees so it points in the opposite direction. This is a 6 meter change.

So would you experience the same amount of Gs if moving in a straight line either vertically or horizontally (assuming both have the same rate of acceleration change)?

No you would not.
For example, if you feel a G-load of 10 G vertical (in the "positive" direction, i.e. the direction associated with upward acceleration), then you are accelerating a little more slowly than if you feel a G-load of 10 G in a horizontal direction.  Or to put it another way, if the magnitude of the actual acceleration is the same in both cases, the "felt" G-loading is higher when the object is accelerating vertically upward than when the object is accelerating horizontally.  For example, if the actual rate of acceleration is 10 X 9.8 m/s/s, the resulting net G-loading is 11 G if the object is accelerating vertically upward, while it is only 10.05 G  (the vector sum of 10 x 9.8 m/s/s horizontal acceleration and 1 x 9.8 m/s/s vertical acceleration) if the object is accelerating horizontally.1
To understand why, please read on -- we'll try to make a bit of sense of all this!
(And please note that this apparent "discrepancy" between G-loading and actual acceleration, when we are dealing with a vertical acceleration, works in the opposite direction from what would be needed to help explain your intuition that the G-loading "felt" by astronauts during a rocket launch is abnormally low, compared to the G-loading felt by the pilot of a maneuvering  jet fighter!)
G-loading is related to acceleration, but it is not exactly the same as acceleration.2  G-loading is the vector sum of a body's (e.g. a vehicle's) actual acceleration vector, minus the 1- G earthward acceleration vector caused by gravity.
Here's another way to say that -- G-loading is what we get when we take the vector sum of all the forces acting on a body other than gravity, and then divide by weight. In the context of flight, this means that G-loading is what we get when we take the vector sum of all the aerodynamic and thrust forces acting on a body, and then divide by weight.
If an astronaut feels a G-loading of 3 G on launch, he is only accelerating upward at 2 x 9.8 m/s/s , not 3 x 9.8 m/s/s.  The third G is used to cancel gravity, or if you prefer to put it the other way, the third G is cancelled by gravity, and thus does not contribute to the actual acceleration of the body.  The 3 G figure represents the net aerodynamic and thrust forces acting on the vehicle, divided by weight, while the actual net acceleration, due to the vector sum of all forces acting on the vehicle (including gravity), is only 2 G.
Likewise in straight and level constant-speed flight-- or at rest on the ground-- the crew (and the G-meter) "feel" 1 G, but their actual acceleration due to the vector sum of all forces (including gravity) is zero.  For the in-flight case, the 1-G figure is due to the upward lift generated by the wings.  Since thrust and drag are equal and opposite, this also represents the net vector sum of all the aerodynamic and thrust forces acting on the aircraft.  At rest on the ground, the 1-G figure represents the upward force of the earth on the tires, transmitted to the bodies of the occupants through the landing gear, aircraft structure, and seat cushions.
The fundamental reason3 for this somewhat odd state of affairs is that unlike aerodynamic and thrust forces, which act externally on a body, gravity "acts from within".  It exerts an equal acceleration on every molecule of a body, and thus creates no stresses or strains of any kind4, and so can't be "felt", and can't be detected by a G-meter.5  It has no tendency to pull a pilot down into the seat or sideways against the side of an aircraft.  That's the secret to understanding weightlessness.  In the arcing trajectories flown by the famous "Vomit Comet" astronaut training aircraft, aerodynamic lift is reduced to zero, and thrust is set to be equal (and opposite) to drag, and therefore the net vector sum of all aerodynamic and thrust forces is zero.  Gravity is still acting on (accelerating) the aircraft and crew, but gravity can't be "felt", so weightlessness results, and the occupants float freely around the aircraft, and the G-meter6 reads zero.
Likewise in actual space flight-- beyond the atmosphere, aerodynamic forces are always zero, so the G-load is always zero (and weightless always exists) whenever the engines are not producing thrust.7  In this situation a G-meter will always read zero-- which does not mean that the vehicle has travelled beyond the reach of the earth's gravitational field!
Disclaimer: all this is nuance.  It would be sufficient to simply point out that the very high G-loadings (e.g. much more than 4) sometimes experienced by jet fighter pilots are always acting in the vertical (up-and-down) plane in the aircraft's reference frame, and are due to strong lift forces produced by the wings, not due to high thrust forces produced by the engines. Even a catapult launch off an aircraft carrier only averages around 3 G's of acceleration.  Adjusting the numbers up or down by 1 G, to properly account for the role of gravity, doesn't change the basic conclusion here.
It is also relevant to point out that even a modest (linear) acceleration, sustained over a long time period, will eventually produce a very high velocity (speed).  That's what explains the very high speeds reached by rockets, despite their relatively modest acceleration rates.8
Footnotes:

In this paragraph-- and throughout this answer unless stated otherwise-- we are talking about objects on or near the surface of the earth, and thus experiencing the normal gravitational pull of the earth that we feel at the surface.

"G's" of acceleration are not the same as G-loading. G's of acceleration are simply equal to the acceleration divided by 9.8 m/s/s. If a vehicle is accelerating purely horizontally, at a rate of (for example) 1 G, this does not mean that the actual total G-loading is only 1 G. If we say a vehicle is accelerating upward at a rate of 1 G, it's arguably a bit unclear whether we really mean that the actual net acceleration is 1 G, or we actually mean that the G-loading is one G, in which case (for an object in the earth's gravity field relatively near the surface) the actual net acceleration would be zero.  The former appears to be the more correct interpretation of what is meant by "accelerating upward at a rate of 1 G."

We're speaking from a Newtonian perspective here.  The relativistic perspective on gravity is different, and is beyond the scope of this answer.

Other than tidal effects, which are normally only significant on the planetary scale.  Tidal effects are due to one side of a body being in a stronger gravity field than the other side of the body.

Obviously, given certain constraints, a G-meter will indicate the strength of the gravitational field acting on a body. For example, if the body is known to be at rest on the ground, or is known to be travelling at a constant speed in a constant direction.

In this answer we're referring to a hypothetical G-meter that measures acceleration in all three dimensions.  The actual G-meters found on aircraft instrument panels only measure acceleration (excluding the acceleration component caused by gravity) in one dimension, the "up and down" direction in the aircraft's own reference frame.  This is normally the component that is most likely to become dangerously large in a poorly executed maneuver, turbulence, etc, due to the capability of the wings to generate very high lift forces, if allowed to meet the air at too high an angle-of-attack in relation to the airspeed.  In flight, a G-meter that only registers accelerations acting in the up-and-down direction in the aircraft's own reference frame, is essentially simply deflecting in proportion to the lift force generated by the wing (at least so long as the actual weight of the aircraft is held constant.) That's why a G-meter is an essential tool for aerobatic pilots who wish to avoid accidentally pulling the wings off the airplane.

Note that this is true even when the trajectory is completely linear, e.g. a spacecraft travelling directly toward the earth, or directly toward the sun.  It's not necessary to introduce the fictitious pseudoforce called "centrifugal force", associated with a curving flight path, to explain weightlessness, regardless of whether we are talking about the arcing trajectory of the "Vomit Comet", an orbiting spacecraft, a stunt plane temporarily "unloaded" to zero G's while inverted at the top of an (improperly flown?) loop, etc.

"Relatively modest" meaning "compared to the max G's pulled by a fighter pilot in a high-speed, high-lift maneuver"!

<< at rest on the ground-- the crew (and the G+meter) "feel" 1 G >>
and
<< gravity "acts from within"... and so can't be "felt", and can't be detected by a G-meter. >>
These are contrasting, which is the correct one? 1G or 0G?

Understand, according to the text in your answer the g-meter would be 1 on ground, correct?

Ok, but then I don't understand why you also write that gravity cannot be detected by a g-meter...

So if I jump up in the air I do not fall back to earth, the earth actually rises up to meet me?

MichaelHall -- I never actually said that  (but you've heard the old expression "may the road rise to meet you"--

It was a snarky response to the remark about feeling the earth pushing up against you.  While technically correct on some level, I have always felt such distinctions detract from, rather than enhance our understanding of the way things work.  And may the road rise up to meet you too!


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Who approves flight plans for repetitive helicopter rides at local festivals in the US?
The helicopters are taking off and landing in a public park located in the midst of a residential area.

The question, as asked, is "nobody". Flight plans don't need to be approved by anyone. Flight plans are filed by pilots (or, for big commercial or charter operations, by the planning department of the company) solely in order to tell ATC where the pilot plans to fly.*
(If they're within the controlled airspace of a nearby airport, they'd need to contact that airport over the radio for permission every time they took off, but I don't think that's exactly what you're asking.)
For the specific helicopters you're talking about, though, its highly likely that there is no flight plan at all. You don't need to file a flight plan to fly in the US, so if all they're doing is taking some passengers sightseeing/joyriding, I doubt they'd bother.
Now, if the park is inside the city limits, there may be some noise ordnance or zoning rule that would need to be set aside, but answering that is beyond the scope of aviation.se. You'd need to contact the city council/board/alderman/whatever to ask about that one.
*Well, IFR flight plans can be modified by ATC when they're activated, which I guess could count as a form of "approval" depending on your point of view.

You touched on this a little... to be precise, the flight plan does not need to be approved by anyone, but the clearance is whatever ATC gives you. From that moment on the "plan" is all but irrelevant. Also note that (at least in the USA) a filed VFR flight plan is not helpful in telling ATC where you plan to fly, because VFR flight plans are not viewable by ATC.

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Can airplanes rear wheels spin on take off (like a car launching from a halt, where its rear tyres spin before they get grip)
Just wondering (especially in the rain) if an airplanes rear wheels will spin like a cars rear tyres will spin when it accelerates from a halt.

The wheels are not powered like a car. They provide no motive force.

Ok. Sorry if it sounds like a stupid question im not a huge aviation guy lol, appreciate the answer tho thank you :)

If an airplane was on a treadmill...

@MichaelHall -  Too funny

I believe there have been experiments in driving the main gear wheels in the past, either to help planes accelerate at the start of TO, or to give the ability to taxi without actual engine thrust, but the benefits weren't worth the weight/complexity. Keep in mind that weight is Priority One in design when it comes to airplanes, and extreme measures are often taken to keep weight down, like using exotic materials that no car designer in his right mind would use, or making seemingly trivial choices between fastener A or B based on some microscopic weight difference that adds up to a few pounds.

@JohnK I think in most cases this would be hard to accomplish.  The more common use case I’m familiar with is spinning up the landing gear before landing to reduce tire wear.  It was tried a few times and in each case, the cost of the extra fuel to fly the drive mechanism around ended up being more than what is saved on the tire budget…  so it’s never made it’s way onto any mass produced aircraft that I’m aware of.  The drive mechanism required to drive an airliner around, for instance, would be very large and very heavy.

No, the tires on a plane aren't driven by the engine. All the power comes from the propeller/turbine pushing on the air.
Well, actually, there have been a (very) few experimental or one-off planes with driven wheels. But even then, the vast majority of them had wheels that were driven by low-power electric motors without enough torque to actually "burn rubber". Those motors wouldn't have been used for takeoff, just as a more fuel-efficient way to taxi around the airport.
If the brakes are dragging, it is entirely possible for it to skid "the other way": i.e. with the engine pulling the airframe forward while the tire doesn't rotate. But that's about it.

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Why is the maneuvering point (Va) in the V-n diagram defined as the upper intersection point? Why not the lower point one?

in my textbook and apparently everywhere online I have found that Va or cornering velocity is defined to be the point of intersection where the stall line and the limit load factor line meet. Apparently, it also states that this Va is the lowest airspeed at which load factor can be reached and any speed below that cannot exceed the limit load factor because the airplane will stall first.
Ok, I see that. If I so horizontally to the left from Va (around 120kts) yes, the plane will stall first. however, it seems to be that this speed is not the lowest airspeed to reach that load factor. isn't the lowest airspeed around 80kts? which is at the bottom part of the intersection around when negative Gs (between -1 and -2gs)are experienced.

Because it is unlikely that you fly upside down for an extended period.
You are right that it is possible to exceed the negative limit by aerodynamic means below v$_A$ but that does not mean that the structure will fail immediately. Normally, the maximum positive and negative structural load factors are about equal; the biggest exception occurs with wooden spars (wood is stronger in tension than in compression) or if buckling limits the compression load of the spar. Note that aerobatic airplanes typically have the same value of load factor in both directions.
The low negative limit is used because more is not needed and some things become easier with a lower load factor, especially for the airplane systems (engine, instruments, controls).
The next reason are gust load factors. Starting from 1 in level flight, an upward gust will cause a load factor 2 gs higher than a negative gust of the same magnitude. Since it is assumed that you fly right side up in gusty weather, the lower limit can be 2 gs smaller (in absolute terms) than the upper limit.

I'm not sure I follow your reasoning (at 80kt, the aircraft would stall at either $\approx$ 1.6g or $\approx$-1.3g, both smaller than the limit load factors +3.8g and -1.5g), but it's just a definition. You could define a "negative manoeuvring speed" and take the negative limit load, but the consensus for manoeuvring speed is taking the positive one.
While flying, we usually experience more positive load factors (e.g. turns, flaring out,...) than negative ones. Therefore, I am not surprised to see the positive load being used for the definition.

Because under normal flight conditions (for non-acrobatic activities), we fly at or around 1 G. So deviations caused by updrafts, down drafts turbulence, etc. that might cause airframe overstress are much more likely to occur in the positive G direction than in the negative G direction.
Since "maneuvering speed" in General Aviation is defined in order to give pilots an idea as to what speed puts them at risk of overstressing the airframe due to these deviations, it is the positive G inflection point that is most relevant.
But you are right, if a string enough downdraft occurred (near a thunderstorm or in heavy turbulence, for example), it could cause an overstress if you are faster than the indicated/calibrated airspeed at the lower negative G inflections point.


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why does the rotor could head (Nr) "overspeed" if the collective is lowered suddenly, what is the Physics behind it
I know that in helicopters the N2 and Nr are coupled. I would appreciate a physical explanation leading to the cause of how the rotor blades could overspeed if collective (power) is lowered suddenly
I have heard that the rotor head needs to have a torque load in order not to overspeed. but I am not sure how that works.
The only thought I have in mind is if I am spinning something and suddenly I stop doing so, it is going to spin at the speed it had when I stopped the motion and/or less and decrease from there. But I dont see how can something could "overspeed" if the motion (energy) is stopped. The kinetic energy will be dissipated by lowering its speed but it cant increase, if so how and why? thank you

I think you should see this too: https://aviation.stackexchange.com/q/62029

On a turbine helicopter, the engine is being managed by a governor that maintains the rotor RPM (that you trim to a specific value in the green arc). When you move the collective, you are changing the torque demand on the engine and the governor senses the resulting incipient rotor RPM change up or down, and lowers or raises engine torque by increasing for decreasing the speed of the gas generator (the gas generator in a free-turbine turboprop or turboshaft is equivalent to the engine core on a turbofan).
There is some lag to this response loop. If you drop collective really fast while the engine is producing significant torque, there's a delay because the governor doesn't respond by reducing torque until the RPM starts to increase, the response itself has a bit of a lag as an internal control loop, and unloading the blades by lowering collective provides less resistance to the existing unreduced torque.
So basically if you slam the collective down fast enough, the engine gets unloaded faster than the governor can respond by reducing engine torque, and you might get a momentary overspeed until the governor catches up, especially if the RPM was near the top of the green arc when you did it.

thank you, "less resistance to the existing unreduced torque" Are we talking about the resistance/drag caused by the high AOA on the rotor blades hitting the relative wind towards the head??

Yes. The blade is just an airplane's wing forced to go in a circle. Shove the nose down on an airplane with a high power setting and keep the power on until you actually see the airspeed start to rise before you cut it back. You'll gain a few knots to the extent the surplus thrust remained in place. The reactions are just way faster with a helicopter rotor. It's the opposite of cutting off power and delaying lowering collective for a second. RPM decays. Just like delaying lowering the nose in an airplane, after the power is cut, drops the speed.

The engine keeps pushing the rotors just as hard (at least at first), but the drag goes down so the rotors speed up.
You've got the metaphor backwards. Imagine instead that you are removing a bolt with a wrench. It's slow going and takes a lot of effort. If the wrench slips off the bolt but you're still pushing hard on the wrench, what's going to happen?
When you notice it start slipping, you'll stop pushing so hard. But not quickly enough that the wrench doesn't turn pretty fast at first. This is the same with the helicopter- the governor will work to reduce the torque and keep the rotor RPM the same. But if the collective is dropped too quickly the rotor will speed up at first.

awesome thank you, "but the drag goes down so the rotors speed up" Are we talking about the drag caused by the high AOA on the rotor blades? and when the collective is lowered the AOA on the rotor blades reduces creating less drag?

@YamchaAviator Yes.

In a 701D Engine for the UH60M Blackhawk there is no over speed.  The load demand input from the collective schedules and meters fuel but if the rotor speeds up the DEC (Electronic Control Unit) will fine tune the fuel and air via a torque motor.  This motor also controls the air equation with inlet and variable geometry (vanes stationed between the Stage 1&2  compressors.  Simply put, the fuel/air mixture.  The ECU does this just like modern cars.  If there is a malfunction then you may get an overspeed.  It has droop response as well.  More collective equal droop but the torque motor controlled by the EDECU (newest version) will via milli-seconds react.  Overspeed is controlled by an overspeed circuit.  Trims the fuel then sent to the injectors.  Sensors (Np, Q, Overspeed) at the rear power turbine reports this to the EDECU where the process continues.  The N1 turbine (gas) is not coupled to the N2. Gases simply at 51k RPM kinectic energy move the Np turbine at almost 21k RPM.  This separate shaft turns the high speed shaft at the same speed which is connected to input module which ultimately moves the rotor at 258 RPM (gearing). Other helicopters may a work a bit different. Only thing to add is that the Q Motor does while the LVDT senses the result and sends the signal back to the EDECU.  Fine tunes over and over.    Hope this helps.

What is the source of this text, please?


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what is the lower pressure plate in the swashplate assembly? is it the same as the lower stationary swash plate? or is it the drive scissors?
Im looking at the rotor assembly of a BlackHawk. The natops manual states the following, "...The lower pressure plate, in conjunction with the main shaft nut, secures the shaft extension to the main shaft. The lower pressure plate also provides an attachment for the swashplate assembly. The swashplate has stationary and rotating discs separated by a bearing. It transmits flight control movement to the main rotor through four pitch control rods. The swashplate slides around the main rotor shaft and tilts in the direction of the flight controls on a Teflon-coated uniball."
What exactly is the lower pressure plate? based on the drawing it looks like its pointing to what in other books/sources its called "drive scissors" (the articulable 2-piece link attached to the main shaft to the rotating plate)? or is it the rotating upper plate the lower pressure plate? thank you in advance.
I just don't see any other plate than the fixed and rotating plate..
last, what is the main shaft nut? Is it part or under the fairing or is it part of the fairing?

"The lower pressure plate, in conjunction with the main shaft nut, secures the shaft extension to the main shaft. The lower pressure plate also provides an attachment for the swashplate assembly" what's not clear in the explanation? How the extension of the shaft works? Or how the scissors works?


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Can I assume that aoa on horizontal stabilizers correspond to aoa of the main wing?
I am in the process of designing a small wing (Re ~ 50-100k) and am wondering if given (a) the Cm on the main wing is relatively small at low aoas, (b) the horizontal stabilizer will produce lift as long as it is not at 0° aoa and (c) it has some amount of leverage over the main wing, will trimming the horizontal stabilizer to some aoa result the same amount of aoa in the main wing?
There will probably be some discrepancy introduced by Cm on the main wing not being zero, but I feel like that might be small enough to neglect?

Depending on the Reynolds number and the general dynamics over the history of the flow, the wing will produce a wake that will generally change the angle of attack (AoA) observed by the horizontal stabilizer (compared to the AOA of the wing from the freestream). This is quite case-dependent depending on the airfoil shape, as larger camber implies larger downwash of the flow in the wake, and is difficult to predict. The point of the horizontal tail is to stabilize the aircraft with elevator deflection, and the drag generated by it is usually negligible compared to drag from other components.

The question is in the Vote To Close queue. It has been flagged as Needs Details Or Clarity. Yet when I read the question, I fully understand it. Voting to keep open.

If the wing center of pressure is perfectly positioned over the center of gravity, the tail will rotate around the center of gravity until it is in a zero lift condition.  Normally, CG is set slightly forward of wing CP, requiring some tail downforce.
Therefore the tail is doing two jobs in straight and level flight, maintaining wing AOA and keeping the net sum of all pitch torque forces (including any off-set thrust vectors) at 0. This is generally accomplished on an as needed basis with trim tabs once proper elevator deflection is established by the pilot.
What makes it a bit more complicated is that wing CP can change with AoA, generally moving forward and increasing with higher AoA.  This is can partially compensated by deliberately placing the horizontal stabilizer in the downwash zone of the wing, providing additional counter balancing tail force.  The Cessna 172 is an excellent example of this subtly brilliant design feature, allowing for use of a strong lifting cambered wing.
But downwash effects should not be "assumed" or "neglected", they should be carefully studied (wind tunnels), tested, and quantified; at high angles of attack control authority of the aircraft can be most important.  Downwash can result in a significantly different AoA on the tail than one might expect from the free stream.  One can modify or eliminate these effects by changing the vertical placement of the horizontal stabilizer on the tail.

"wing CP can change with AoA", "placing the horizontal stabilizer in the downwash zone of the wing" ... uh ... I think I have some homework to do :-)

@fho Not to worry.  Because of much lower wing loading, classic airplane designs scale down to models well.  Just remember tail must control the wing.  With models, a few crashes along the way will help build your knowledge (in a much safer way).  Regarding wing loading, lower speed of models plus Reynolds number effects need to be considered.  The more experience you get building and flying the better you will be.  You may wish to start with a similar sized model from a hobby store that flies within your planned speed envelope.

I guess "iterate until perfection" is something that is really daunting as a newcomer. Actually this is already Mk2. Mk1 was basically build without any knowledge of aerodynamics and works ... ok. Now it's time to get a somewhat solid grasp on the theory before startin Mk2.

After, oh, a million years, given the same set of conditions, intelligent design and trial/error will arrive at the same solution, with one faster and the other well tested.  Sometimes it is a little bit of both.

I will note that one of the reasons behind the T-tail was to get the hstab out of the wings downwash-affected area and into the free stream because it simplifies the analysis considerably. The downside is that in a deep stall the hstab can move into the roiling air behind the stalled wing reducing elevator effectiveness in some cases making recovery impossible.

Yet many sailplane designs successfully use T tails.  Huge swept wing airliners need stick shakers.  And you can err slightly large with your model H stab.  You may wish to give it a go.

The short answer is "No" the AoA of the horizontal stabilizer is not the same as the AoA of the wing.  For the aircraft to remain in the air the AoA of the wing must be positive.  if a positive AoA were seen by the Hstab, positive lift would also be produced by the tail, but a downforce is required from the tail in order to stabilize the aicraft.  Therefore the AoA of the stab must be negative.
This, by the way, leads to the idea of the canard where the forward stabilizer needs to produce a lifting force rather than a downforce to stabilize the aircraft.
Even if the pitching moment in level flight from the wing/body were zero, then there stabilizing force from the tail would need to be zero as well. So, again, positive AoA on the wing, zero AoA on the stab.

Welcome to Av.SE!

@RalphJ - Thanks.  I’ve resisted joining for years (‘cuz I really don’t need another time sink :-) but here I am…


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What is a "suction airfoil"?
I just stumbled upon this entry on airfoiltools.com:

http://airfoiltools.com/airfoil/details?airfoil=griffith30symsuction-il
There is basically no information there and (a quick cursory) web search did not yield anything useful. My search bubble seems to focus on airfoils that actively suck in air to prevent flow separation, which this design apparently does not do?

I think your search bubble is correct. The airfoil profile in the question does, however, defeat the purpose of suction airfoils, as their function is based on suction provided to the upper latter part of the airfoil, accelerating the flow along the upper surface and keeping the flow attached behind the suction point by removing the turbulent boundary layer:
Design, Analysis, and Evaluation of a Propulsive Wing Concept - Gavin K. Ananda (Researchgate)

(link to picture on Researchgate site)
A symmetrical airfoil of this design would not provide any benefit as the acceleration of airflow would be present on both sides of the airflow, pretty much negating the effect in any AoA.
The symmetrical form in question does remind me of the Synergy aircraft fuselage profile. Synergy claimed on it's previous, now remodeled website that the fuselage form profile provided positive thrust under certain conditions, assisted by suction provided by the pusher configuration propeller. There is much debate available regarding these claims on the internet, just google "synergy aircraft pressure thrust".


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Why do white marks on the spinners in front of engines have different patterns? Not Yellow? Omitted for some models?
In aeroailas' answer I understand that the cones in front of the props/fans are called spinners.
In Farhan's answer I understand that the ehite marks painted onto the spinners act less as scarecrows and more plausibly as scare-crews.
In Daniel K's answer I understand that the marks may have yellow color for high contrast.
Then I found a picture from Wikimedia showing different patterns for the white marks:

View Original Picture
Courtesy of Gese 151, shared under CC BY-SA 4.0 license.
There are also engines lacking such patterns, as shown for this Jetblue Embracer 190

View original picture
Courtesy of John murphy, shared under CC BY-SA 2.0 licence
Here are my questions:

Why to the airliners have such variety of patterns? What are the pros and cons of such patterns? Why are some patterns less common?
Why don't they opt for other colors in practise for more contracst, like how British use to paint their trains yellow?
Why do some models omit them? Is the change reducing costs? Is such reduction surpassing potential risks?

I doubt anybody can offer an irrefutable answer with the level of detail you seek.  Remember there are many companies in many different country operating these planes, local custom, and pure random decisions probably have as much to do with it as anything.  Like how highway signs are different from country to country, yet they are still effective.

To be clear, my answer on that other question about yellow paint was regarding a test of an experimental prototype engine.  As far as I know all spinner patterns in production engines are white.


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Why does Harrier have split hot exhaust nozzles?
This is a repost after the original questions were deemed too complicated.

Hawker Sea Hawk shot by Smudge 9000 under CC-SA 2.0 licence.
Harriers seem to inherit elements from Sea Hawk, that the exhaust from a single engine is diverted to two side exhausts.
@Peter Kämpf answered my previous question that Harrier was basically built upon Bristol/RR Pegasus engine, which drew inspiration from Michel Wibault's Gyroptere VTOL assembly.

From "S.A. WIBAULT FRENCH-GERMAN SOUCOUPE 'GYROTER'?" (1944) By Rob Arndt

Layout of the Pegasus. Public domain.
However, I noticed the contrast between two designs. Gyroter is more of a shaft-driven lifter like (X)F-35, as contrasted to direct-lift of Harriers. The hot exhaust of Gyroter directly exits at the aft in a single circular nozzle. The cold exhaust nozzles could have further forward positions than Harriers, allowing the placement of the core engine to the further aft, instead of closely around CoM, while maintaining thrust balance.
@Carlo Felicione , in his answer to Why does the Harrier jet have four landing gears? , said that the two aft nozzles forced Harriers to adapt bicycle landing gears.

Close-up of P.1127 Kestrel (Harrier prototype) shot by Alan Wilson under CC-SA 2.0 licence.
You can see the limited space for rear landing gear bay. Also the width of the tail assembly is directly limited by the separation of rear nozzles. It needs extra heat protection as well.
Also in the question Is it possible to modify an AV-8B Harrier II to a supersonic aircraft by applying some modification? @Aeroailas has shown the BS100 engine, designed for the Harriers' supersonic successor P.1154,  adding aftburners to the cold bypasses (PCB) for supersonic flight.

Source: aeromodelbasic.blogspot.com
And I found P.1216, a improved decendant of P.1154. In Wikipedia the change is explained:

The problem with using PCB was that its thrust developed considerably more heat, noise and thrust induced vibrations compared with that produced in the Harrier. As a result, the rear fuselage configuration employed on the Harrier was not suitable. Beginning in 1980 the design team began investigating possible solutions. By removing as much of the rear of the airframe as proposed in the preceding P.1212 and P.1214 designs they developed a small short fuselage with twin booms which removed the tailplanes from being from being directly in the path of the engine's exhaust. A similar solution to the three-nozzle problem was adopted in the later YAK-141 prototype supersonic VTOL interceptor.

The configuration allowed for two nozzles at the front and a single 90-degree rotating rear exhaust which was more efficient than having to split it into the twin rear nozzles used on the Harrier. The booms provided a low drag location for the carrying of cannons in the front as well as missiles and avionics, while also shielding the jet exhausts from heat seeking missiles. Instead of the Pegasus engine whose design dated back to the 1950s the design used the Rolls-Royce RB.422.48 with PCB fitted only on the front nozzles. The design allowed the engine to be removed without needing to remove the wing as on the Harrier.

Source: Buttler, Tony (2017). British Secret Projects : Jet Fighters since 1950 ( 2nd edition) (Hardback). Manchester: Crecy Publishing. ISBN 978-1-910-80905-1.

BAE's ASTOVL concepts. Public domain.
So here's my question: Why does Pegasus engine have two widely seperated aft (hot) nozzles instead of one straight TVN? Why Harriers place their tail assembly between such nozzles?
Wouldn't it be better to have aft exhausts:

At bottom-sides in the further aft like Yak-38?

Uploaded by Tosaka under CC-A 3.0 licence.

Between two tail booms like P.1216, Yak-43/141 and (X)F-35?

Archived in Roy Cochrun Collection

Sheathed in the exhaust duct like X-32?

Power system of Boeing X-32, archived by globalsecurity.org.

Or elevate the tail assembly like Phantoms?

F-4K using British Spey turbofans. Public domain.
Any of them will eliminate the hurdles of strength requirement, heat protection, bicycle landing gears and thrust loss.

The big problem faced by VTOL is getting the thrust-to-weight ratio above unity. So anything that reduces thrust or adds weight is bad. The P.1127/Harrier, being the first practicable solution, wwas uncomfortably close to the borderline.
Long tailpipes waste energy and reduce thrust, so it had to be short. But a long rear fuselage then gets in the way. De Havilland's solution with the conventional Vampire had been to split the tail, and this would one day also be adopted for the Hawker P.1216 VTOL project and today's F-35 Lightning II. But Sidney Camm at Hawker worried at the time that it would add too much weight. So his solution for the Hawker Sea Hawk had been to split the tailpipe. It had worked very well, perhaps better than expected. Meanwhile a complex multi-ring system for thrust vectoring, such as used on the F-35, was untried and posed a significant technological risk to the project. The split nozzles would need only a single swivelling joint each, much reducing the risk. So he wanted to keep them for his VTOL design.
The layout also had the advantage of a "four-poster" powered lift airstream, which helped to make the hover more stable and controllable - another vital feature in those days of VTOL pioneering ignorance. Rolls-Royce built their "flying bedstead" free-flying hover test rig to develop the necessary control systems.
It all worked. The P.1127 became the world's first practical VTOL jet.

Would it be possible to have a tail highly above the tailpipe?

@CrystallizedRefresher Possible but unwise. It makes for increased frontal area and the risk of deep stall.

Why does Harrier have split hot exhaust nozzles?

Because a central tail exhaust is too far from the two front exhausts which are fed by bypass air. The split exhaust makes all ducting mercifully short, so the airframe can be lightweight and compact. Don't forget that pressurized gas cannot be ducted in air conditioning pipes; it needs heavy pressure pipes to do so. As I said in this answer, the P1127 was barely able to hover with 700 lbs of equipment stripped and fuel for 3 minutes. A long and heavy duct for pressurized gas was completely out of the question.
Regarding the reheat of the P1154: This was always a pipedream and the airplane would had melted its tires well before it would had been in the air. Next, fireproofing the fuselage structure would had required a titanium belly and screens of pressurized air to limit structural heating. This was demonstrated by the VJ-101 X2 (which had afterburners in its podded engines) where heat loads in hover with reheat were unmanageable.
A central tail exhaust is only feasible with lots of lift close to the fuselage tip. Note that the F-35B creates more lift with its forward fan than with its rear nozzle. But this kind of complexity is necessary if reheat is to be used for supersonic flight. And using two engines will carry the risk of an unrecoverable roll on engine failure. When the engine fails in a single VTOL jet during hover, the pilot can still safely eject. Not so if two engines are used.


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I need some help to understand the rationales behind some STOVL jet designs, especially the Pegasus-Harrier configuration
In the question:Why do VTOL aircraft (F-35 or Harrier) have only one engine? Aeroailas answered why mature STOVL fighter designs only use one main engine, but did not answer why there are no mature twin-engine STOVL fighter designs.
Also in the question Is it possible to modify an AV-8B Harrier II to a supersonic aircraft by applying some modification? He has shown the BS100 engine,  adding an aftburner to plenum chamber for supersonic flight.
Carlo Felicione, in his answer to Why does the Harrier jet have four landing gears? , said that the two aft nozzles forced Harriers to adapt bicycle landing gears.

Close-up of P.1127 Kestrel (Harrier prototype) shot by Alan Wilson under CC-SA 2.0 licence.

Layout of the Pegasus. Public domain.

Power system of Boeing X-32, archived by globalsecurity.org.
The only other mature STOVL system utilizing direct lift is that from X-32.

Hawker Sea Hawk shot by Smudge 9000 under CC-SA 2.0 licence.
Harriers seem to inherit elements from Sea Hawk, that the exhaust from a single engine is diverted to two side exhausts.
However, now I am even more confused by the design. The following are my questions.

Why does Pegasus engine have two widely seperated aft (hot) nozzles instead of one straight TVN? Even if the thrust loss is negligible, isn't the extra need for strength and heat management increasing weight?

Cont. to Q1, why Harriers place their tail assembly between such nozzles? From the pic above, anything in that tail has to be protected by extra panels, and the width of the it is limited by the separation of the nozzles. Wouldn't it be better to have aft exhausts at side-bottom in the further aft like Yak-38? Between two tail booms like P.1216, Yak-141 and (X)F-35? Sheath the exhaust in a duct like X-32? Or elevate the tail assembly like Phantoms? Any of them will eliminate the need for bicycle landing gears.

Again, why are the two bow (cold) nozzles so widely separated? The air is forced through 90° bypasses after leaving LP compressor, and another 90° before exit. What would be the thrust loss and strength requirement?

Cont. to Q3, why are the two bow nozzles in the same height of aft nozzles? As I see from the pic, the cold exhaust directly blows onto the aft nozzles (or their farings). If it is fine for cold exhaust to zig-zag through the bypasses, wouldn't it be better to travel smoother bypasses (like 135°-135° bents) to the side-bottom, at least to a lower height than the aft nozzles?

Cont. to Q4, or is it possible to just bleed cold exhaust into one TVN at the bottom of fuselage? (The Harriers' gun and ammo pods are located at two sides of the fuselage, so they shouldn't be a hurdle.)

Cont. to Q3, when in level flight, why not retract the bow nozzles, and let cold exhaust flow into a straight bypass around engine core, and mix with hot exhaust to be expelled from the aft? This will streamline the fuselage to better suit supersonic flight. Also this will ease the placement of aftburners, rather than in the cramped plenum chambers.

Why X-32 still uses two awkward-shaped lift nozzles for vertical landing, while it has (unlike Harriers) plenty of space as well as heat-proof linings in the exhaust duct to transform into a single, central, downward TVN?

Harriers have their Pegasi in their CoM to balance when hovering, making two pairs of nozzles rather close to CoM and to each other. Is it possible to have larger separations? E.g. Place engine core and aft nozzle to further aft, and LP compressor and bow nozzles to further bow, only to be connected with a central shaft, so that the extra space around the shaft can be streamlined and holds more fuel and ordnance.

What are the blades in the nozzles called? Are they necessary, as they aren't seen in later generations' designs?

Is it possible to go shaftless in transmitting power from the core engine to the front lift component in any STOVL configurations (as proposed in McDonnell-Douglas' JSF)?

Is it possible to have twin-engine STOVL jets? E.g. combine two Pegasi to build a heavy fighter, somewhat a supersonic Do-31.

that's a lot of questions in one. please consider splitting this in multiple posts so that the questions can be answered properly without needing to write a book.

@Federico Would splitting questions into many posts be considered as spamming?

not if made in a sensible way. you can post a question every x days instead of all at once, for example

How about a six-engine supersonic VTOL jet?

Second prototype of the VJ 101 in hovering flight (picture source).
By using swivelling wingtip-mounted engines, a twin-engine STOVL jet is feasible and it will be much easier to swivel the whole engine, so avoiding the complexity and efficiency losses of sviveling exhausts. The downside is twice the likelihood of an engine failure (which, when happening in hover, will translate into an immediate and unrecoverable roll) plus a much higher roll inertia which results in reduced agility. Wingtip engines were chosen in the hope that the hot exhaust from their afterburning version would be far enough from the landing gear.
Another VTOL jet with more than one engine was the Yakovlev 36 which flew first in 1963. It had two fuselage-mounted engines because no single one at that time was powerful enough to lift an airframe with the desired specifications. Since engine failure in hover meant unrecoverable roll, the series version Yakovlev 38 used a single engine, the development of which had started with the Yak 36. Like the Harrier, it used jet exhausts situated at the sides of the fuselage since a single exhaust would be placed too far away from the center of gravity. Like the Harrier, it could only fly at subsonic speeds because an afterburner in combination with mid-fuselage nozzles is impossible.

Yak 36 in flight (picture source)
The Kestrel/Harrier concept started with the most efficient engine at its core, adding the remaining parts of the airframe later. This meant a high bypass turbofan (which explains the width of the engine and the spacing of the exhausts). Using two engines of half the size would incur a weight penalty. Remember that the P1127 prototype was only just capable of hovering, with fuel for only 3 minutes in the tanks and 700 lbs of equipment stripped for the first hover tests.
The Pegasus engine started with Wibault's Gyroptère concept, a brochure of which he left with USAF colonel Chapman in Paris in 1956 who subsequently got Stanley Hooker interested in this idea. This meant a VTOL-capable engine was already running in 1959, so the P1127 / Kestrel had the basis for its design already laid out when development started.
The reason why the VJ 101 was never developed into a mature design is politics, plain and simple. While NATO doctrine in the late Fifties and early Sixtes demanded VTOL airplanes (including the planes for supplying forward bases, which explains the Do-31), this was abandoned in the mid-Sixties, leaving the engineers and their prototypes stranded in the middle of the development program. By that time only the Harrier was far enough into its development to make a continuation to a series airplane viable.
I'm sure you will understand that I consider a single answer sufficient to your 11 questions.

You explained that: 1. It is better to have single large engine instead of two smaller engines; 2. The spacing of cold exhausts; 3. Kestrel/Harrier is built around Pegasus; 4. Other different designs were cut midways. Yet, I believe some questions remain unanswered.

Pegasus was inspired by Gyroptere, but Gyroptere had a different principle, closer to (X)F-35 lift fan config. It doesn't answer my questions about: 1. Zig-zagged cold bypass, as the blowers were not fed by front intakes; 2. Split hot exhaust nozzles, as it has a straight one to the aft; 3. The placement of Pagasus around the CoM, as Gyroptere can have lift components widely separated and linked by shafts.

I have been avoiding mentioning tilt-engine VTOLs as they are more engineeringly nightmarish. Even wet, the six turbojets of VJ101 cannot surpass one turbofan of Harrier in thrust, and their max VT weight are not comparable. What I asked was a heavy fighter equiped with two Pegasus-like engines and having weight and thrust comparable with (Spey) Phantom.

@CrystallizedRefresher Nobody in their right mind would ever use two Pegasus engines if it can be avoided by any means. The Do-31 had no other chance, but fighters would be limited by the largest available engine. They can be twice the weight of a Harrier if this engine is an F-119.


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wingtip vortex in negative angle of attack
If a wing is at a positive angle of attack, then the vortices produced at the wingtip will generate downwash and induced drag
but on the contrary, if the wing is at a negative angle of attack, then the vortices produced at wingtip will generate upwash and induced thrust
so in this case ( negative angle of attack ) , are the wing tip vortices still harmful or not ?

Just look at that wing with positive AoA upside-down: et voilà

Sophit is right. The only difference the -ve AoA will make is that the vortices will rotate in the opposite direction. The wing will still produce induced drag just as it would at +ve AoA.

Vortices do not produce up- or downwash, they are the consequence of it. Up- and downwash are the result of pressure differences around the wing. You might also read that up- and downwash are caused by circulation (which is an abstract concept from potential theory). This is correct but less intuitive to understand.
Next, we need to agree what Angle of attack means. This could be the geometric angle between the direction of movement and a reference line on the wing (mostly the chord of the airfoil), or the difference between the actual direction of motion and the one where no lift is created. For your question the latter definition will be much more useful.
At positive angle of attack the wing creates lift, so it will deflect the oncoming air downward. To fill the void that the downward movement would leave behind, air will flow into that space from the sides, causing a rotating motion. Inertia will keep that motion going until viscosity damps it down.
If the wing has washout, the inner wing might create positive lift while the outer wing creates a downforce. Any lift on the wing, regardless of direction, will cause induced drag. Downforce at the tips will, however, lower the bending moment that the wing is subject to and allow to build the wing lighter, thus saving induced drag indirectly by lowering the requirement for lift. Note that at the tail and in the downwash of a lift-producing wing, indeed the pressure distribution which causes the downforce can also create induced thrust.
Wingtip vortices are neither beneficial nor harmful. They are only an unavoidable byproduct of lift creation with limited wingspan.


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load factor in high altitudes
if load factor equation is lift devided by weight , so in very high altitude for example 1000 km above the earth atmosphere, there is no air to produce lift so load factor is zero but in newton equation we see that even for very high altitudes there is a small gravitational accelereation due to object weight not lift! so how load factor is same as g but in high altitude load factor is zero but object feels gravtiy?

In most cases the centripetal force is small enough to be ignored, although in supersonic flight it would be measurable.  However, in retrograde (east to west) flight the aircraft is revolving more slowly than the Earth’s surface and so would in fact appear to be heavier than when at rest.  Load factor doesn’t generally make allowance for the direction of flight, which would need to be taken into account for flights at near-orbital speed.

At an altitude of 1000 km, gravity is not so weak (g=7,3 m/s2), and the atmosphere is extremely thin, so there can be no lift at all, and the 'load factor' has no meaning. It's true that satellites 'fly' orbiting planet Earth, at very high altitudes, but that's a different thing, those satellites being in continuous free fall, so the Earth's gravitational acceleration, as measured in the satellite, is zero...

how gravititional acceleration is zero and object is in continuous falling?

Imagine yourself in an elevator hanging from a rope. Suddenly, the rope breaks and –assuming no air drag and other resistances– the elevator and its contents fall at 9,8 m/s2. As you are falling –and accelerating– with the elevator, you are weightless... The acceleration that you feel or measure with a dynamometer is zero...

Just to add to the answer of @xxavier over here: the object (satellite) is not only falling like an elevator with the brocken rope but it is falling on a curved path which is just perfectly circular around the earth $\Rightarrow$ it is in continuous free fall. And that's why people on the ISS are weightless and not because they are so far from earth that they do not feel gravity anymore

@alireza: You are confusing load factor and Gs which are two different things. Load factor is a pure number (lift/weight), while G measures acceleration i.e. length/time². Being one of the two zero does not imply that the other one is zero too.

@xxavier Dynamometer? did you mean accelerometer?

An accelerometer is essentially a dynamometer that measures the inertial force that appears when a test mass is subjected to an acceleration...

Here, like in your other question, we have a typical situation when definitions break down when removed from their context. All definitions, especially engineering ones, have a certain domain (often implied) where they remain valid and meaningful.
What is lift? Do engines in a vertical climb provide lift, or what? Is load factor the thing that the G-meter in the cockpit shows? Does it have to point down? What is "down"? What is weight?
1. Weight
In a purely physical sense, you can define load factor as the ratio between the weight and any reaction force that counteracts this weight. We don't need to call this force "lift". When you are standing on a static floor, the floor provides a reaction force that is exactly equal to your weight, and you have "load factor" 1. Is it enough? Is it useful?
This depends on how you define "weight". Surprisingly, there is no universal definition: weight is not an essential physical quantity. Mass is. In many cases, weight is defined exactly as the reaction force, i.e. what the scales show. If they show 1000 N1 when you are standing on them on the floor, then that's your weight.
Now let's build a 1000 km tower (with all the amenities) and put you up there. At that elevation, gravity is less (but not too much less), and your scales will show about 750 N. Now that's your weight! The load factor is still 1 by our definition: the floor reaction ("lift") matches your weight.
Now let's put you in a centrifuge and spin you up. The scales under you might show 2000 N now. Still, the floor is intact, and now that's your weight - even though in this case not all of it comes from gravity! And the load factor still 1.
Doesn't feel very useful, right? But this comes from the "common" definition of weight: you "actually" feel lighter in the first case and heavier in the second, and the floor is stressed less on the tower and more in the centrifuge. If you buy cheese on the tower "by weight", you'll get more of it for the same money!
So, it seems more useful to define "weight" in the load factor equation as some standard quantity, fixed for the "normal" gravity at the surface. In this case we'll relate the actual reaction force (lift or whatever) to your "normal" weight of 1000 N. And then we'll get load factor G=0.75 on the tower and G=2 in the centrifuge. And this actually reflects your stress, as well as the stress of the floor (or airframe).
Note that the G-meter in an aircraft is essentially a weight scale, calibrated to the ground level and graded in "g" of 9.81 m/s2.
Stop! But what about the "lift = weight" formula for the level flight? It doesn't work with such "fixed" definition of weight. Yes, there is different weight here! The one we used in the first definition.
This is the crux of the problem. In our daily life, these weights are "practically" the same, and we conflate them. But once you step out of the "normal" conditions, like putting the aircraft in outer space, the implicit conditions in each definitions may no longer be valid, and you need to watch carefully what you mean by each term.
2. Lift
What is lift? Is it the force that "lifts" the body away from Earth and holds it from falling? Or is it the "useful" force that the wing generates? This term starts to break down when the attitude is such that the two directions don't align (like in a vertical climb), or when there is no aerodynamics to speak of.
From the point of view of the aircraft (and its engineer), lift is the aerodynamic force normal to the incoming stream flow.2 Most of the attention is given to this force, and generally to the forces in this direction which is "vertical" for the airplane: they are by far the greatest, and the most stressful for the airplane. The typical G-meter also measures the load factor in this plane. But nothing prevents us from measuring the forces (and defining the load factor) in other planes: after all, pilot's body (and the airframe) experiences stress in all directions.
In a steady vertical climb, aerodynamic lift is zero, but its role is taken up by the engines. The load factor is 1 (when close to ground), but it's in a different direction, and it's not the direction in which the normal load factors are certified. Still, you (and the airframe) will feel it; you just might not care that much from the point of view of stress (or might care more, if it was a side force). But from the point of view of physics, there is no fundamental difference.
In outer space, we have no aerodynamics. But we can still provide "lift" by other means. When a jump jet takes off vertically, its engine arguably provides lift, while its wing is useless. So, get a rocket engine, direct it "down" (towards gravity), and throttle it to provide just enough thrust to keep the altitude. And you'll have a non-zero load factor. Which one? At 1000 km, a G-sensor will show about 0.75, as we discussed earlier. The direction depends on attitude.
When you run out of fuel (which will happen in minutes, if not seconds), you'll start falling down. There is no other force you can use to counteract gravity. So you'll have zero load factor, like in any free-fall situation. Ironically, you don't "feel" gravity when gravity is the only force acting on you.
To avoid crashing into Earth, you can move forward such that you "miss" Earth due to its curvature. That's how satellites fly. But you are still falling constantly. This is why they don't feel weight in orbit.

1 Weight is, of course, force. But common scales are usually (and mistakenly) graded in the units of mass (e.g. kg). Even if we consider it be the related force unit such as "kg-f", this has an implicit definition of the gravity acceleration "g", which is important for our discussion. So let's stick to the proper (even if uncommon) units.
2 And in the plane of symmetry, to separate it from side force.

thanks a lot bro. so in vertical climb g is 1 not zero because in this case lift is produced by engines not wings? based on newtons second law if an aircraft goes vertically up without any accelereation the pilot and aiframe feels 1g . based on load factor point of view , in vertical climb there is also lift but not produced by wings and it is duty of thrust to produced the same lift. is my conclusion correct?

@alireza: in a vertical climb at constant speed, G would be 1 longitudinally to the aircraft (from nose to tail) and G=0 vertically (from belly to canopy). Also lateral G (from left to right) would be 0. Load factor is by definition lift/weight = 0/weight = 0. Engine produces a thrust to win weight plus aerodynamic drag. This should be the complete view about this case.

@sophit, right, except that here again, it's better to qualify the "definition" of load factor. In this "normal" definition, it is implied that this is the "body vertical" load factor. Since it is usually the main point of concern, it is the "default" one. But the longitudinal load factor (which is 1 here) also exists; it's just rarely a concern and is generally well within ±1 G even for very powerful jets, so it's rarely mentioned or certified. Yet when we are trying to work out the confusion in non-usual situations, it's better to qualify things explicitly.

@Zeus: yes sure, I meant the standard definition of load factor, the one that you find in the books i.e. lift/weight. Very good answer.


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is aircraft weightless during vertical climb?
in vertical climb , if velocity is constant (no acceleration) the pilot feels 1g gravitational acceleration that is equals for crusing flight. but in vertical climb lift must be zero so load factor is zero that means 0g
which one is correct and why?

Lift isn't zero, it just isn't being provided by the wings.

@RonBeyer in vertical climb lift must be zero otherwise rather than verticaly going upward , the aircraft make a loop

Lift is zero, thrust equals drag+weight, load factor is zero, aircraft would be weightless in free fall

so what is the answer? in vertical climb g is zero according to load factor or 1?

Do feel weightless in an elevator going up?

In a vertical climb, "lift" comes from the engine, not the wings.

Having been in an F-16 going vertical from 12k feet to ~17k feet, at a constant speed, I can assure you that were were not "weightless or "0g". My back against the seat was a constant 1g.

@RonBeyer: lift is zero in this case, by definition. Thrust against drag and weight is the right term.

@alireza: You are confusing load factor and Gs which are two different things. Load factor is a pure number (lift/weight), while G is measures acceleration i.e. length/time². Being one of the two zero does not imply that the other one is zero too. Maybe this answer can help.

While established in a steady climb or descent, the plane is in equilibrium. Hence the 1g. It’s similar to being in straight-and-level: plane’s not weightless in any of those situations. The plane too feels 1g. If the plane were to be weightless, the pilot would be too. @sophit makes a good point

No, the aircraft is not weightless.
In level flight the pilot feels 1 G in the vertical axis, through the seat of the pants.  The wings also "feel" 1 G because they are supporting the weight of the aircraft.
In pure vertical unaccelerated flight what the pilot feels is rotated 90 degrees to the rear, and he/she will feel 1 G against the backrest of the seat.  The wings also "feel" 1 G to the rear, but they are unloaded and not supporting the weight of the aircraft as they do in level flight so you could say the load factor is zero.
The wings are not generating any aerodynamic lift, they are acting as simple fins, like on a rocket in vertical flight.  The load is borne by the engines.

but in vertical climb lift must be zero so load factor is zero that
means 0g

Here is where you are going astray.
The best definition of the G-loading is the net force acting on the aircraft, minus the force of gravity, divided by (mass * G).1
In other words, the G-loading is the net "felt" force acting on the aircraft, divided by (mass * G).
In other words, the G-loading is the "felt" acceleration component, attributable to the "felt" forces acting on the aircraft.  Gravity is not a "felt" force, and does not contribute to the "felt" acceleration.2. That's why we can find the net "felt" force by subtracting away the force of gravity from the net force.
In flight, the only forces acting on the aircraft other than gravity are the aerodynamic forces generated by the aircraft, and thrust.3  So in flight, the G-loading is the vector sum of the aerodynamic and thrust forces, divided by (mass * G).
Note that (mass * G) is just another expression for weight.
The point we're getting around to here, is that the G-loading may act in any direction in relation to the aircraft.  It is not limited to pointing "upward" or "downward" in the aircraft's reference frame-- though that's the only component that we measure with a traditional panel-mounted mechanical G-meter.
So in a steady-state vertical climb, the G-loading is 1 G skywards in the earth reference frame, and 1 G nose-wards in the aircraft body reference frame.4  Because that is the vector sum of Thrust and Drag, divided by Weight.  Or speaking in terms of scalar values, we can say that Thrust minus Drag = Weight.
Once we recognize that the G-loading can act in any direction relative to the aircraft, we have to abandon any conception of a fixed relationship between G-loading and wing loading.  When we talk about a fixed relationship between G-loading and wing loading, we are really only talking about the component of the G-loading that acts parallel to the lift vector.  This is, basically, the component of the G-loading that is measured by a traditional mechanical G-meter mounted on the aircraft instrument panel, or by a 1-axis electronic G-meter mounted in the same orientation.5  Because the maximum lift force that can be generated by the wings dwarfs any other aerodynamic force acting on the aircraft, and also dwarfs the thrust force, this is the normally only component of the G-loading that is of any real practical concern to the pilot, or to the aircraft designer.6  But in the broader aerospace context, we need to treat the G-loading as a vector that may point in any direction relative to the aircraft.
Footnotes:

This is vector arithmetic.  We can't just work with the magnitudes of the force vectors, unless they are lined all lined up to point in parallel (same or opposite) directions.  Example: steady state vertical climb at constant airspeed, net force is zero, so if we subtract the 1G downward pull of gravity, we get a net G-loading of 1G upward.  See footnote 4 for more on the sign convention at play here.

The reason that gravity cannot be "felt"-- or detected by the G-meter or the slip-skid ball-- is that the acceleration component due to gravity creates no stresses or strains on the aircraft structure, or on the contents of the aircraft, because gravity "acts from within", creating an equal acceleration on every molecule of the aircraft and occupants.

Strictly speaking, we should also consider apparent "centrifugal forces" arising from the rotation of the aircraft about its own CG.  For example, if the aircraft is in an extreme flat spin with a very high yaw rotation rate, the longitudinal component of the G-loading, as measured by a fore-and-aft oriented accelerometer, would be different (opposite in direction) at the front and the rear of the aircraft.  In most cases, this effect is negligible.  Don't confuse this with the component of the G-loading that is associated with the curvature in the flight path, the curvature in the path followed by the CG of the aircraft.  The latter is already fully accounted for, when we compute the net force acting on the aircraft.

Some may argue that the direction of the G-load is reversed in this paragraph.
That would be taking the viewpoint that when we stand at rest on the earth's surface, the stresses and strains we "feel" in our bones and tendons and muscles is due to the downward pull of gravity, so the G-loading should be considered to be 1 G downward.  Actually, when we stand at rest on the earth's surface, the stresses and strains we "feel" in our bones and tendons and muscles is due to the upward push of the ground against our feet.  Nothing would change in what we "feel" if gravity suddenly vanished, but the ground kept pushing up against our feet as usual (which would require that the ground suddenly be rapidly accelerating upwards.)
Anyway, if we want to choose to adopt the convention that the +1 G load we feel when at rest on the earth's surface acts in the downward direction, then we have to recognize that we are really talking about an apparent load that is the mirror image of the actual "felt" component of the net force acting on a body, divided by (mass * G).

"Basically", because the lift vector does not always act exactly parallel to the body z axis of the airplane, i.e. parallel to the direction of "up" in the aircraft's own reference frame.

Except, of course, during a catapult launch -- or a crash!

Related ASE answers:
Does lift equal weight in a climb?
What is the definition of load factor and how do you apply it?
Aircraft load factor and body normal acceleration


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effect of wing chord length on lift coefficient
In the bertin's Aerodynamics book
It is written that reducing the wing chord increases the lift coefficient( decreases stall angle) and as a result, the wingtip stalls earlier than the wing root
how ?

Does this answer your question? How would you adjust C_l airfoil values for different Reynolds numbers?

What do you mean by reducing wing chord? Keep the same overall area and increasing the aspect ratio of the wing, OR
Keep the same wing span and change chord length so that also the wing area is smaller

We know the definition of the aircraft lift coefficient in steady flight:
$$C_L=\frac{W}{\frac{1}{2}\rho V^2 \cdot S}$$
We can now divide the entire wing (with area $S$) in tiny slices of area $dS$, each with width $dy$ and chord $c(y)$.
$$ dS = c(y) \cdot dy$$
We can substitute the above equation to get the local lift equation:
$$C_l(y)=\frac{ dW}{ \frac{1}{2}\rho V^2 \cdot c(y) \text{ } dy} $$
What happens to the to the local $C_l(y)$ when the chord is low or when it's high?
How does $c$ vary for a typical wing from chord to root?

For a typical tapered wing, with an elliptical lift distribution the lift and the local lift distribution look like this:

As you can see the outboard lift section is highest, meaning that section will stall first if all sections have the same definition.
Source: E. OBERT, Aerodynamic design of transport aircraft (IOS Press, 2009).

thank you but what is happening in phisiycal point of view ? i mean when chord lenght reduced , reynolds number also reduced wich means lower stall angle and lower lift coefficient ! can you tell me what exactly happens in terms of reynolds number ?

The question that sophit refers to explains how Cl values change for varying Reynolds numbers

Cl and stall angle both decreases with decreasing reynolds number but we know that higher Cl means stall earlier ! how can get higher Cl and lower stall angle with reduction in reynolds?

Local stall happens when the required $C_l$ ( $C_{l_{req}}$ ) exceeds the $C_l$ the airfoil can deliver ($C_{l_{max}}$). Decreasing the value of $c$ drives up the value of $C_{l_{req}}$ due to the reasons I mentioned above as well as reducing the value of $C_{l_{max}}$ due to the reynolds effects

@alireza: $C_l$ does not decrease with Reynolds number rather with the AoA. $C_{l_{max}}$ decreases (and $C_d$ increases) at smaller Reynolds numbers.


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Can a sonic boom occur in supersonic Mach number?
If the wing's leading edge has a sufficient swept angle , then it is called a subsonic leading edge.
Therefore, if the air speed is supersonic, the air speed on the wing will still be subsonic.
According to the mentioned cases, can it be said that normal shock can be produced on wing  in supersonic mach number as well? And the second question is, is it possible to break the sound barrier in supersonic Mach due to the fact that the wing's leading edge is subsonic and in higher supersonic much numbers the flow reach mach 1 on the wing surface ?

This seems to confuse terminology with what's actually happening on the wing. From this source (bottom of page 4), "If the wing leading edge is swept inside the Mach cone the component of Ma perpendicular to the leading edge is subsonic; hence, the swept wing is said to have a subsonic leading edge. .... A swept wing with a subsonic leading edge behaves somewhat as a wing at subsonic speeds, although the actual free-stream Mach number is supersonic."

if the airflow over the wing is subsonic, then the aircraft is transonic.

Therefore, if the air speed is supersonic, the air speed on the wing will still be subsonic

Not really
Only behind a normal shock wave the flow becomes subsonic. Behind an oblique shock wave the component of speed perpendicular to the wave is subsonic but in total the flow is still supersonic, a bit slower but still supersonic. Slide 7 of this presentation gives a good overview of this:

The blue arrow is the flow before the oblique shock wave while the red arrow is the flow behind it. No matter what, $u_2$ is always subsonic, while $w_2$ always equals $w_1$ and the red arrow is always supersonic.
If we make at supersonic speed the same assumption like at subsonic speed that the lift characteristics of a wing do not depend on what happens along the span of the wing but only perpendicularly to it, then having a sweep angle bigger than the oblique shock wave makes the wing "subsonic". But only in respect to its lift characteristics: the flow on the wing is still supersonic.

But the formation of the shock wave depends on how fast the air hits the wing's leading edge vertically. Even one of the ways to delay and reduce the power of the shock wave is the use of swept back wings.
Therefore, the vertical component of Mach number that hitting the wing leading edge is not the only determinant of the amount of lift production.
So can it be said that normal shock and as a result breaking the sound barrier can occur at supersonic speed?

A sonic boom occurs whereever plane/object speed exceeds the speed of sound. Since speed of sound unlike light is based on variable criteria (temperature, pressure, density) average is 343 meters per second so any object not Painstakingly aerodynamic will generate some kind of boom

The speed of light is also based on variable criteria.


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induced drag at supersonic speed
for a delta wing with supersonic leading edge at small angle of attack there is no stagnation point on leading edge so There is no connection between the top and bottom of the wing. Therefore, the air cannot move from under to top of the wing .
does it means zero induced drag ? i mean leading edge vortex and wingtip vortex

This should answer your question.


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Why is the wave drag of a straight wing less than that of a swept back wing?
According to NASA's claims, when the swept back wing is inside the shock cone and the shock is not attached to the leading edge of the wing(subsonic leading edge), the wave drag of the swept wing is less than that of a straight wing, but with increasing speed and decreasing the angle of the shock , the shock attached to the leading edge of the swept wing(supersonic leading edge), so in this situation the wave drag of straight wing is less  than the wave drag of the  swept back wing! Why ?
i mean wave drag(not total drag )

First, you should always provide a citation when you say something like "According to NASA's claims".  It will provide the context that we can make sure we are answering the right question.
I actually think there is some confusion here depending on how formal people are being.  In any case, we can probably look at this by just considering a 2D airfoil.
The transonic drag on an airfoil will often be higher than the supersonic drag.
In the supersonic case, the shock is in front of the airfoil (like the wing behind the shock case you describe) and the airfoil exists entirely in the post-shock flow.
In the transonic case, the shocks will lie on the surface of the airfoil -- likely both the top and bottom surface.  There, the shocks cause a strong adverse pressure gradient on the airfoil surface (pressure jump).  This is very likely to cause the boundary layer to separate.  This separation causes a large increment in drag.
This is the cause of additional transonic drag.  It isn't really wave drag, but it is separation drag caused by the shockwave.
The same thing happens in 3D.  When the shock is in front of the wing, the wing exists in relatively benign post-shock flow.  When the shock is on the wing surface, the wing sees a strong adverse pressure gradient at the location of the shock -- which causes separation and lots of drag.


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What is the speed and temperature profile of the boundary layer at the stagnation point?
At the statgnation point where the air velocity is zero, what is the shape of the velocity and temperature profile of the boundary layer? Is the velocity profile of the boundary layer in the form of a straight line (because the air velocity is zero at all points)?
what about temperature profile ?

By definition the boundary layer begins at the the stagnation point, so its thickness there is simply zero. Or do you mean something else?

The shape of the velocity and temperature profile of the boundary layer is going to be a flat line(parallel to the surface). This is because at the stagnation point, the air will stop its movement, which means no velocity or speed around the points of the boundary layer, thus creating a flat line. The same can be implemented for the temperature profile. When the air stops moving, the movement responsible for creating different temperatures stops and the temperature around the boundary layer becomes uniform and becomes parallel to the surface.

Could you rephrase this more clearly?  "The temperature ... becomes parallel to the surface"?  A temperature gradient can happen without convection (moving fluid), too: conduction and even radiation still happen.

For Mr Camille Goudeseune: Thanks for the comment, you can rephrase it as "The temperature profile around the boundary layer... becomes parallel to the surface."

Hi A.C.A., here's how this works on stackexchange.  Comments like the one I added suggest how to improve something.  Then whoever wrote the original thing clicks on Edit, and makes the improvement.  That's better for future readers than making them read through a long conversation.


text:
why lift coefficient decreases at supersonic flow?
In Anderson's performance book, he wrote that the higher the speed, the greater the pressure difference between two points, and as a result, the lift coefficient is greater.
But when we reach the speed of sound, due to the interference of the shock wave and the boundary layer and the separation of the boundary layer, the lift coefficient decreases.
My question is: in supersonic speed, the speed behind the shock is still supersonic, so the pressure difference between two points in supersonic speed is more than subsonic speed, and the separation of the boundary layer is less than the separation that exists in the speed of sound(transonic speed).
So why does the lift coefficient decrease at supersonic speed?

"in supersonic speed, the speed behind the shock is still supersonic" Not necessarily: the component of speed perpendicular to the shock is always subsonic, so the total speed after it might actually be subsonic

but we always use M2 that is superconic and not Mn2 that is subsonic !

Could it be that you mean to ask why the lift curve slope decreases in supersonic flow?

@PeterKämpf yeah

As you probably know already, the 2D lift curve slope in supersonic flow is $$C_{L\alpha} = \frac{4\cdot\alpha}{\sqrt{Ma^2 - 1}}$$ when calculated by potential flow theory. But I guess you would like to understand better why that is.
In supersonic flow the only zone of influence which a wing can have is within its Mach cone. With increasing Mach number this cone gets narrower, so less air can be "reached" by the wing in order to be accelerated downwards and to create lift. This makes the wing less efficient the higher its Mach number becomes.
This might be a very simplified view, but it captures the essential mechanism well. At the same air density, a higher Mach number also means a higher dynamic pressure, so even as the coefficient of lift decreases for the same angle of attack at higher Mach numbers, the absolute lift will still grow.

Let's consider a flat plate and let it fly both at subsonic and at supersonic speeds.

Let's start with the subsonic case and let's see how the airflow looks like around the flat plate (picture source)

At subsonic speed the airflow has enough time to move out of the way and "get around" the flat plate. As depicted in the picture, the airflow starts to bend well in front of the flat plate and basically all around and far from it. The bending of the streamlines generates a relevant pressure field (according to Bernoulli) which is always globally pointing upward i.e. it generates a total aerodynamic force which is always a pure lift. The fact that at subsonic speed no drag is generated around an aerodynamic body is called D'Alembert paradox.
Drag obviously does exist at subsonic speed and is mainly due to the fluid's viscosity but as a first approximation we can simply ignore it and live with the fact that the airflow "encircles" the aerodynamic body (our flat plate) generating a pure lift.

What happens now at the flat plate at supersonic speed? The flow is now so fast that it simply cannot realise the presence of the flat plate until it just crashes against its leading edge. At that point the flow suddenly changes its curvature to follow the flat plate's surface. And when it reaches the leading edge, it just changes (suddenly again) its curvature to become again parallel to the freestream (picture source):

Also at supersonic speed this sudden change in speed comes with a relevant change in pressure (lower on the upper surface; higher on the bottom) but this time the pressure acts everywhere perpendicular to the flat plate's surface i.e. it always possesses a component pointing backward: at supersonic speed, even considering an inviscid fluid, the aerodynamic force (R in the previous picture) is always composed by a lift and by a drag. This latter is called wave drag.
For that reason alone the same flat plate flying at supersonic speed generates a lower $C_l$ than at subsonic speed. Anyway at supersonic speed there's much more going on, with changes in density and temperature which cannot be neglected and which globally contributes to increasing dissipation and to reducing lift in respect to the subsonic case.


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what is acceleration of gravity in free fall situation with air drag?
consider a ballistic missile in reentry trajectory with initial velocity of 1000 m/s. as missile enters the atmosphere, air drag decelereting its velocity so missile's acceleration becomes negative. in this situation how many g's i mean acceleration due to gravity missile felt? positive? negative or zero g? is there any eqution for that?
another question is when air drag is equal to acceleration of gravity  is it means that g is one?
sorry about my english

Good question, but in my opinion it belongs to Physics Stack Exchange.

I’m voting to close this question because it's about physics, not aviation.

Just to compare apples to apples, Force = mass × acceleration = mass × $delta$ Velocity/time
Drag is also a force.  This makes it easy, because forces can be added and subtracted.
Before entering the atmosphere, there is no drag, so downward force is mass × 9.8 m/s$^2$.  G force felt is 0.  Free fall.  Velocity is increasing.
As drag starts, Net downward force is (mass × gravity) - drag force, still positive, but less.  G force is greater than 0 but less than 1.  The object continues to speed up, at a lower rate.
Downward acceleration stops when (mass × gravity) - drag force = 0
This is terminal (constant) velocity. G force felt = 1 towards earth.
If you have ever watched a Falcon 9 launch, terminal velocity decreases as the booster falls into thicker atmosphere.  Now acceleration is negative, and G force is greater than 1.  The rocket is slowing down.

Minor correction: acceleration is not v/t, it is ∆v/∆t. Both are very different.

I guess that's what we're trying to show as net force.  Positive delta v is less as drag kicks in, eventually going to zero, then negative as the rocket slows down in thicker atmosphere.

If you accelerate from 1000 m/s to 1002 m/s in one second, what is the acceleration? "v" (average v) is 1001 m/s, ∆v (change in v) is 2 m/s. By using v like you did, we get an acceleration of 1001 m/s², by using ∆v like I suggest, the acceleration we get is 2 m/s². Clearly, the latter is the actual acceleration. I get your point, and for your case, the more appropriate term would be dv/dt (instantaneous acceleration) instead of ∆v/∆t. But v/t is wrong because it doesn't represent acceleration, as shown in my little calculation above.

@AdityaSharma please feel free to edit in the deltas.  I agree this is a more accurate way of describing acceleration.

It won't let me edit, it says "edits must be atleast 6 characters long" , you can yourself just add a delta before velocity and time.

Search "delta symbol" on web and simply copy-paste it. Wait! you can copy pase it from here: ∆ ,there you go. LoL!


text:
What, exactly, is a final controller?
This http://www.xavius.com/ordapp.html tutorial mentions the user work song as a final controller for two runways and says that during busy periods one controller would work each runway.  I’ve never heard of this before and couldn’t find anting by googling on my own.  So, what is a final controller and are they tower or TRACON?

The final position exists as described below, but you will almost never hear US ATC call the position as such when speaking to pilots. Outside the US you will hear it much more often (almost always?) if the position exists.

Final controller isn't a What, Final Controller is a Who.  Air Traffic Controllers are people.

"Final" or "Final vector" is the position in charge of, well, final approach. It is a radar control position worked in the TRACON.
As aircraft are transferred from the ARTCC to the Tracon, the Center (or Enroute) controller will tell them to contact the "Approach" controller. Everyone in the TRACON uses the callsign "Approach" (or "Departure"),* but the first person the pilot talks to in the TRACON will be the arrival position. This position is responsible for initial descent and sequencing of several arrival streams coming in to the TRACON's airspace.
After the arriving aircraft have been descended and slowed somewhat, they are transferred to the final position. This position is responsible for actually turning aircraft onto final approach course in sequence and with proper separation, and for monitoring separation along final approach to make sure the planes don't get too close. A full "PTAC"—position, turn, altitude, clearance—is quite a mouthful to say and eats up a good bit of frequency time, so during times of heavy traffic it is beneficial for there to be a dedicated controller for every one or two final approach courses; if they were also responsible for initial descent and sequencing they wouldn't have time to get out everything they needed to say. Having a dedicated controller at a dedicated radar scope also allows the scope to be zoomed in more, which increases detail and can allow for a tighter sequence because the controller is more confident that what they think is proper 3NM separation actually is.

*There is a very different type of radar controller who actually talks planes down to the runway using radar vectors only, instead of vectoring them onto an instrument approach and having the pilot follow a radio signal being radiated from the ground for guidance—this is an ASR or PAR approach. This position actually uses the callsign "XXX Final Controller" instead of "XXX Approach." Not all TRACONs have ASR/PAR approaches available; it requires dedicated training, and the controllers have to be sure to do so many per quarter to maintain currency with them, and for PAR approaches it requires specialized radar equipment. They are more common at military bases.

In addition to the above answers, the term final controller can also refer to the controller issuing instruction for a ground controlled approach, such as a PAR or ASR approach. Aircraft will get handed off to this controller before commencing the approach, and will receive lateral and/or vertical guidance depending on the equipment available for that runway. GCAs are almost exclusively limited to military airfields but used to be more common in civilian use.


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Does there exists on-time data for the EU, like there does for the US?
I've recently found a fantastic dataset straight from the US government. It has detailed data including scheduled departure/arrival time vs actual departure/arrival time, taxi time, cancellation, for every single domestic flight in the US.
The EU is generally good about collecting data like this but I have not for the life of me been able to find it. I was wondering if anything similar exists for the EU.

If such dataset exists, it will be aggregated at Eurocontrol level (each country in the UE has still a somehow independent ATC, with partial integration into Eurocontrol). Eurocontrol area corresponds more or less with the EU territory. Many datasets are also available at the EU Open Data Portal